Chapter 3: Electricity

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Q
Electricity

Question:

Show how you would connect three resistors, each of resistance 6 Ω, so that the combination has a resistance of (i) 9 Ω, (ii) 4 Ω.

Answer:

(i) For 9 Ω

We will connect two resistors R_1 and R_2  in parallel and the third resistor R_3 in the series with the parallel combination of R_1 and R_2.

The equivalent resistance of the parallel combination is given by,

frac{1}{R_P} = frac{1}{R_1} + frac{1}{R_2}

frac{1}{R_P} = frac{1}{6} + frac{1}{6}

frac{1}{R_P} = frac{1+1}{6}

frac{1}{R_P} = frac{2}{6}

frac{1}{R_P} = frac{1}{3}

R_P = 3 Omega

Now, the equivalent resistance of the circuit,
R_{eq} = R_P + R_3

R_{eq} = 3 + 6 = 9 Omega

 

(ii) For 4 Ω

We will connect two resistors R_1 and R_2 in series combination and the third resistor R_3 in the parallel with the series combination of R_1 and R_2.

The equivalent resistance of the series combination is given by,

R_S = R_1 + R_2

R_S = 6 + 6 = 12 Omega

Now, the equivalent resistance of the circuit, R_{eq} will be

frac{1}{R_{eq}} = frac{1}{R_S} + frac{1}{R_3}

frac{1}{R_{eq}} = frac{1}{12} + frac{1}{6}

frac{1}{R_{eq}} = frac{1 + 2}{12}

frac{1}{R_{eq}} = frac{3}{12}

R_{eq} = frac{12}{3}

R_{eq} = 4 Omega


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